Rev. Rc i Chapter 1

D

b

dx

Strain Stress

AS

X-section

M

ARBAMINCH UNIVERSITY CE-451 INSTITUTE OF TECHNOLOGY RC-I

CHAPTER 2

DESIGN OF BEAMS FOR FLEXURE USING WORKING STRESS DESIGN (WSD) METHODS

2.1 Basic Assumption:

1. A section which is plane before bending remains plane after bending. This implies strains across section are

linearly varying. This is true for most section of flexural member except deep beam where shear deformation is

significant.

2. Beam section behaves elastically when subjected to service load moment. This implies stress in the concrete

varies linearly from zero at neutral axis to a maximum at the extreme fiber.

3. Tensile strength of concrete is ignored. The reinforcement assumed to takes all the tension due to flexure.

4. Perfect bond exist between steel bars and concrete such that no slip occurs. This is possible if adequate

development length of bars and concrete cover are provided.

5. The modular ratio, , may be taken as the nearest whole number (but not less than 6 or more than 15).

In doubly reinforced sections, to consider creep of concrete in compression zone an effective modular ratio of

shall be used to transform compression reinforcement for stress computation.

2.2 Design Equations for Singly Reinforced Rectangular Section

Consider a singly reinforced rectangular section subjected to a service load moment, as shown below.

BY: SILENAT D. 1


a) From the strain diagram, similarity of triangles gives

(1)

In elastic range, applying Hooke’s law, the maximum strain in concrete & strain in steel,

&

And, the ratio of these strains

(2)

By definition, is the modular-ratio, n

Equating Eq(1) and Eq(2), and substituting

(3)

, thus k is an indicator of the neutral axis position.

b) Considering equilibrium of a section

i) For horizontal equilibrium

Substituting and ,

(4)

Let --is known as geometric steel ratio

Then,

Substituting it into Eq.(4 )

With, ,

simplifying, or (5)

BY: SILENAT D. 2


From Eq.(3 ),

Ratio of stresses in steel to concrete, rearranging the above equation

(6)

but, from Eq.(5), and equating with Eq.(6),

Rearranging the following second degree equation in terms of ‘k’ is obtained.

Solving for k,

(7)

ii) The internal couple resulting from internal forces and must equal to the external applied service load

moment. The convenient moment center is taken usually the line of action of the internal forces.

--Taking moment of internal forces about line of action of ,

Substituting & , and simplifying then equation of service load moment resistance of

section is obtained as,

(8a)

Letting be lever-arm ratio for internal forces of section of beam, then service load moment resistance of

section may be written as,

(8b)

Letting be relative bending moment of section of beam, then service load moment resistance of


(8c)

Rearranging Eq.(8c), the effective depth of section required by singly reinforced beam obtained as,

BY: SILENAT D. 3


--In similar manner, taking moment of internal forces about line of action of ,

Substituting & , simplifying equation of service load moment resistance of section is

obtained as,

(9a)

Letting be lever arm ratio for internal forces of section of beam, then service load moment resistance of


(9b)

Rearranging Eq.(9b), the area of tension steel required by beam section is obtained as,

2.3 Type of Singly Reinforced Beam Sections-Based on Modes of Stresses

Depending on the amount of steel used by section, singly reinforced sections are divided into three: Balanced section,

Over-reinforced section and Under-reinforced section.

a) Balanced Section: The most economical section in terms of material usage. In this section, the maximum

stresses in both the reinforcement and the concrete reach simultaneously the respective permissible value.

i.e

From Eq.(3), neutral axis depth ratio of singly reinforced section,

For balanced section, ratio of allowable stresses of steel to concrete is denoted by ‘r’ as,

BY: SILENAT D. 4


Substituting ‘r’ into above equation, the balanced neutral axis depth ratio is obtained as

(10)

From Eq.(5), steel ratio of singly reinforced section,

Substituting ‘r’, the balanced steel ratio is obtained as

(11a)

where r—is ratio of allowable stresses of steel to concrete

Substituting equation of from Eq.(10) in to Eq.(11a), the balanced steel ratio is rewritten as

(11b)

This equation would gives the balanced steel ratio of singly reinforced section in such away that the maximum stresses

developed in steel and concrete when section subjected to service load moment will reach simultaneously the respective

allowable stresses. The corresponding lever-arm ratio and relative bending moment of balanced singly reinforced section

are obtained by

b) Over-reinforced Section if > b: Over-reinforced sections are those that contain more reinforcement

than the balanced one. Hence, as the applied moment is increased, the maximum stress in concrete reaches its permissible

value first; and by the time the stress in reinforcement reaches its permissible stress, the concrete is over stressed.

Therefore, the stresses in concrete and steel for such section are as follow:

--determined from stress diagram using similarity of triangles

The maximum moment of resistance of over-reinforced section is obtained by the following equations in terms of

allowable stress of concrete as,

BY: SILENAT D. 5


--used to determine neutral axis depth ratio

or --used to determine area of tension steel

where

Here, an increase of load produces over stress in concrete earlier than the reinforcement; as a result the concrete crushes

in compression. Such failure is sudden and occurs without warning. For this reasons, over-reinforced section is not

recommended in design.

c) Under-reinforced Section if < b: Under reinforced sections are those that contain less reinforcement

than the balanced one. In such sections, the tensile reinforcement is insufficient to develop the full strength of the

concrete in compression, so that when the reinforcement is fully stressed, the concrete is under-stressed. Therefore, the

stresses in concrete and steel for such section are as follow:

--determined from stress diagram using similarity of triangles

The maximum moment of resistance of under-reinforced section is obtained by the following equations in terms of

allowable stress of steel as,

--used to determine neutral axis depth ratio

or --used to determine area of tension steel

where

Here, failure is more gradual than over-reinforced section. As when steel is over-stressed, the steel yields but is still able

to support the yield stress since steel is a ductile material. Therefore, from both safety and economic point of view, it is

recommended to design section of flexural member as under-reinforced section.

2.4 Control of Deflection

The deflection of structure or part of structure shall not adversely affect the appearance or efficiency of structure or

finishes or partitions. For beams and slabs, the vertical deflection limits may generally be assumed to be satisfied

provided that the minimum depth required by deflection specified by code is maintained.

ACI code provide minimum depth required by beams and one-way slabs in terms of span length as given in table below

can be used as a crude estimate of initial depth to control deflection.

BY: SILENAT D. 6


Table: ACI-code minimum depth of beams and one-way slab to control deflection

Note: For other grades of steel, the value given for is modified by multiplying factor

of .

EBCS-2 provide minimum effective depth, ‘d’ to be used to control deflection is given as,

where --characteristic yield strength of steel in

--effective span length; and for two-way slabs, the shorter span length

--constant as given in table below; and for slabs carrying partition walls likely to crack, shall be taken as

--distance in meter between points of zero moment (for continuous beam, may be taken approximately as 0.7 times

length of span), and, for a cantilever span, twice the length to the face of the support

Table: Values of

Member

Simply

supported

End

spans

Interior

spans

Cantilever

Spans

-Beams 20 24 28 10

-Slabs

a) span ratio, (includes one-way

slabs)

b) span ratio,

25

35

30

40

35

45

12

10

Note: For slabs with intermediate span ratio interpolate linearly.

BY: SILENAT D. 7

Types of Member

Simply

supported

End

spans

Interior

Spans

Cantilever

spans

-Beams or one–way

ribbed slab

S-400MPa

S-300MPa

-One-way solid

slab

S-400MPa

S-300MPa

Strain

AS

a) Doubly reinforced section

M

b) Balanced singlyreinforced section

c) Comp. steel plusExcess tens. steel


2.5 Doubly-Reinforced Rectangular Beam Section

If the section of RC beam is limited in dimension (usually depth), it can not develop the compressive force required to

resist the applied bending moment as singly reinforced section. That is, the applied moment is greater than the balanced

moment capacity of singly reinforced section. For small increase of moment over the balanced one, over-reinforced

section can be used, which is not recommended in design.

A more economical and safe way of designing section in such case is to provide reinforcement in compression zone of

RC section. This section termed as doubly reinforced. Doubly reinforced section can also be used if the required depth of

section of beam as singly reinforced is unacceptable. The purpose of reinforcement in compression zone of RC section is

to assist the concrete in resisting compressive force and to keep the neutral axis at the ideal position ensuring balanced

type failure.

In doubly reinforced beam section, concrete and steel act together to take compression. If both steel and concrete behave

elastically, the stress in compression steel is ‘ ’ (modular ratio) times the concrete stress at the same level.

However, concrete under sustained compressive stress deforms continuously with time due to creep effect and concrete is

also subjected to shrinkage over a period of time. Whereas these time dependent effects do not occur in steel. As RC

beam deforms, even at low loading, there is a continuous transfer of stress from concrete to compression steel. Therefore,

the actual stress in compression steel is larger than that computed on the basis of elastic behavior of materials. i.e

where is concrete stress at the level of compression steel.

To approximate the effect of creep of concrete, ESCP-2/83 code species that an effective modular ratio of ‘

’ is to be used to transform compression reinforcement for stress computation with the stress in

compression reinforcement not to exceed the allowable stress of steel, .

2.6 Design Equation for Doubly-Reinforced Rectangular Section

Consider a doubly reinforced rectangular section subjected to a service load moment, as shown below.

BY: SILENAT D. 8


Two couples method is used to determine the required areas of tension and compression reinforcement by treating

doubly reinforced section into two parts. The total resisting moment is equal to the sum of two resisting couples: one of

which is provided by given cross-section of beam without compression reinforcement with a partial tension steel area,

that balance concrete in compression; and the other by compression steel, and the remainder of tension steel

area, . Thus, the section with compression steel is designed as balanced reinforced section in such away that

compression steel and extra tension steel are proportioned by maintaining the balanced neutral axis depth.

Let --balanced moment capacity of a section if singly reinforced

--excess moment produced by compression steel plus excess tension steel

Then, total moment capacity of doubly reinforced section is,

Balanced moment capacity of a section if singly reinforced and the corresponding area of tension steel balancing the

section are obtained by

&

Excess moment resisted by compression steel plus excess tension steel from couple produced by internal forces

developed in the section,

Rearranging the above equation, internal forces developed in compression steel and excess tension steel are obtained as

Then, area of excess tension steel is obtained as,

Therefore, total area of tension steel required by doubly reinforced section is obtained as,

BY: SILENAT D. 9


From similarity of triangles shown in fig.(b) above, the stress in the concrete at the level of compression reinforcement,

,

Therefore, the stress in compression reinforcement,

Due to the presence of reinforcement in compression zone, there is a loss of concrete area of magnitude, . And, this

will cause a corresponding loss in compression force of .

Therefore, if

If , then

2.7 Flanged Section (T- or L-section) under Flexure

In construction of building structures, the slab is usually supported by a system of beams. If the connection between the

beam and the slab adequately transmit longitudinal shear force, then the beam and slab together may act as a

homogeneous section of T- or L-forms. For loaded beam-slab if subjected to a moment which produces compression at

the top surface, the slab therefore becomes parts of the compression flange of the beam, resulting in a greater zone of

compression and giving a more economical section. For a reinforced concrete beam-slab section, adequate connection

between the beam and the slab is easily provided by casting the section as monolithic, and by extending beam-stirrups

and bent bars up into the slab.

BY: SILENAT D. 10


It is known that the compressive stress caused by flexure in the upper flange decreases as the distance from the web

increases. This is because the shear deformation of flange relieves some of compressive stresses as the element becomes

more remote to the web. Therefore, this makes exact analysis of flanged section of infinite wide-flange complex. In order

to simplify the design of flanged section of infinite wide-flange, it is usual to assume a uniform stress over a reduced

width of flange. This reduced width is known as effective width. Effective width of flange is determined equating forces

on compressive flange due to actual compressive stress on infinite wide-flange with equivalent uniform compressive

stress on reduced width of flange.

The effective width has been found to depend primarily on the type loading, span length, spacing of beams, width of the

web, and the relative thickness of the slab with respect to the total beam depth. For practical design of flanged section,

effective width of flange recommended by codes may be used.

-ACI code prescribes a limit on the effective flange width, ‘ ’ as follows.

a) For interior T-section, effective flange width shall be the smallest of:

Where --is span length of the beam.

--is width of the web

--is thickness of the slab

b) For exterior T-section (L-forms), effective flange width shall be the smallest of:

c) For isolated T-sections, effective flange width shall be

and, also ACI code requires that

-EBCS-2/95 also specifies the effective flange width, ‘ ’ as follows.

a) For symmetrical T-beam, effective flange width shall not exceed the lesser of:

b) For edge beams (L-section), effective flange width shall not exceed the lesser of:

BY: SILENAT D. 11


2.8 Design of Flanged beams for Flexure

Design of flanged beams are made depending on the sign of design moment develop in the section producing either

tension or compression on flange side of beam. If the design moment developed in section produces tension on flange

side, the section is to be designed as if it were rectangular beam of width equal to the width of the web of the section. For

such section, no advantage is gained in using slab as flange of section.

On other hand, if the design moment developed in section produces compression on flange side, the section is to be

designed depending on the position the neutral axis. The position of the neutral axis depends up of the proportions of the

cross-section, the amount of tension steel, and the strength of the materials. If the neutral axis lies in the flange, the

section is to be designed as if it were a rectangular beam of width equal to the effective flange width. When the neutral

axis lies in the web, the section is to be designed as T-beam section.

2.9 Design Equations of T-beam Section -Working Stress (Elastic) Method

Consider a flanged section subjected to a service load moment, as shown below. Assume the neutral axis lies in the

web so that the section is designed as T-beam section.

From geometry of strain diagram and assuming perfect elasticity of both materials, expression for neutral axis depth

ratio, is obtained as:

(*)

Since the compression area provided by the slab is so large (large ), the actual maximum compressive stress in

concrete, will be some unknown fraction of its allowable stress value. Hence, neutral axis depth ratio, has to be

BY: SILENAT D. 12


given in terms of the maximum compressive stress in concrete, that does not related to the allowable stress value,

.

To simplify the derivation of design equations, the compressive stress in the web above the neutral axis is ignored.

Therefore, total compressive force in the flange is equal to:

and, resultant tensile force in steel,

For horizontal equilibrium,

(**)

From Eq.(*)

Substituting this expression of into Eq.(**) to eliminate unit stresses and then gives expression of the neutral axis

depth ratio of T-beam section as:

The distance to the center of compression (centriod of the trapezoidal area of compressive stress) from the upper face of

the beam is:

Then, the lever-arm of the couple formed by the internal tensile and compressive force is:

Substituting and , and solving for , the following expression lever-arm ratio of T-beam section is obtained,

The resisting moments of T-beam section are equal to the product of the lever-arm, of the internal force couple and

the total tension or compression. Hence,

--in terms of steel stress

or --in terms of concrete stress

BY: SILENAT D. 13


Approximate equation for resisting moments can be obtained using the limiting values for lever-arm between internal

forces, and average compressive stress in the flange, as,

--used to determine trail area of tension steel

--used to check maximum stress in concrete

BY: SILENAT D. 14


BY: SILENAT D. 15


BY: SILENAT D. 16


BY: SILENAT D. 17

dx y= 0.8x

StrainParabola-rectangle

stressEquivalent-rectangle

stressx-section


CHAPTER-3

LIMIT STATE DESIGN FOR FLEXURE AND SERVICEABILITY LIMIT STATE

3.1. Basic Assumptions:

Assumption made for determining ultimate resistance of a member for flexure and axial force according to EBCS-2/95

are,

1. A section which is plane before bending remains plane after bending. This implies strains across section are

linearly varying. This is true for most section of flexural member except deep beam where shear deformation is

significant.

2. The reinforcement is subjected to the same variations in the strain as the adjacent concrete. This implies there is no

slip between steel bars and the adjacent concrete. This is possible if adequate development length of bars and

concrete cover are provided.

3. Tensile strength of concrete is ignored. The reinforcement assumed to takes all the tension due to flexure.

4. The maximum compressive stain in concrete when a section complete plastic deformation is taken to be

in bending (simple or compound) in axial compression

5. The maximum tensile strain in the reinforcement is taken to 0.01. This limit assumed to limit crack-width with in

tension zone of section to the acceptable limit.

6. Either idealized parabola-rectangle stress distribution or equivalent rectangle stress distribution for concrete in

compression zone given by code as shown below shall be used in derivation of design equation.

The ultimate resistance of section may be determined using equilibrium of both internal and external forces based on the

stress block obtained from the basic assumption.

BY: SILENAT D. 18

D

b

d

x

Strain

AS

X-sectionParabola-rectangle

Stress block

Equivalent rectangleStress block


3.2. Design Equations for Singly Reinforced Rectangular Section:

Consider a singly reinforced rectangular section subjected to a factored load moment, as shown below.

-Equilibrium of both internal and external forces,

i)

Let --steel ratio of section

Simplifying, depth of neutral axis obtained as,

(1)

ii) Where --lever arm

-taking moment about :

Substituting from Eq.(1),

Simplifying, ultimate moment of resistance of section is obtained as,

(2)

The same equation of ultimate moment of resistance of section can be obtained if moment center is taken at .

-Defining the ultimate moment and relative steel-area using the following dimension-less parameters:

BY: SILENAT D. 19


--relative ultimate moment

And --mechanical reinforcement ratio

Then, neutral-axis depth obtained in Eq.(1) can be written as,

(1a)

Therefore, depth of equivalent stress-block is obtained as,

Writing equation of moment of resistance of section in the form as shown below by rearranging Eq.(2),

Writing the above equation in terms of dimension less parameters,

(2a)

Rearranging Eq.(2a),

Solving for ,

(3)

Therefore, area of tension steel required to resist the ultimate moment, is obtained by taking moment about as,

Where substituting from Eq.(1a) and from Eq.(3)

Rearranging, the required area of tension steel is obtained by,

(4)

3.2.1. Balanced Singly Reinforced Section

BY: SILENAT D. 20


In balanced section, yielding of tension steel and crushing of concrete takes place at same time when the section

complete plastic deformation. That is, the maximum compressive strain in concrete reaches the ultimate strain,

and the strain in tension steel is just yielded, .

From strain distribution, using similarity of triangles,

Substituting & , the balanced neutral-axis depth is obtained as,

(5)

Where --ultimate compressive strain of concrete

Equating with equation of neutral-axis depth obtained in Eq.(1) and Eq.(1a), the balanced reinforcement ratio and the

balanced mechanical reinforcement ratio are obtained as,

(6)

And (7)

If , the steel yields first at the load near collapse (a case of under-reinforced section and ductile-type failure).

If , crushing of concrete takes place first prior to yielding of tension steel at the load near collapse (a case of over-

reinforced section and brittle-type failure).

To ensure ductility, in practice the maximum amount of tension steel is fairly below the amount corresponding to the

balanced-one.

ACI:318 code recommend: maximum reinforcement ratio ensuring ductility as . For seismic load

resisting member, the same code recommends, .

Based on ACI recommendation ( ), maximum design constants of singly reinforced section are obtained

as shown in table below.

Table: Maximum design constants of singly reinforced section (ACI-code)

EBCS:2/95 recommend: the maximum amount of tension steel used to ensure ductility is based on limiting the neutral-

axis depth at,

--for no redistribution of elastic moments

BY: SILENAT D. 21


--for 10% redistribution of elastic moments



Based on EBCS-2/95 recommendation, maximum design constants of singly reinforced section are obtained as shown in

table below.

Table: Maximum design constants of singly reinforced

section (EBCS-2/95 code)

Better approach as follows:

In accordance with LSD method, at ULS of collapse:-

εc approaches εcu = 0.0035

BY: SILENAT D. 22

Steel Grade

S-300S-400S-460

0.4370.4010.382

0.3410.3200.309% Redistribution of

elastic moments

0%

10%

20%

30%

0.3584

0.2944

0.2304

0.1664

0.294

0.251

0.204

0.152


The reinforcing steel shall yield first (s

yd

dy E

f )

Þ Ductility is ensured by means of under reinforcement.

At balanced failure simultaneous failure of the two materials (Concrete & Steel) occurs.

Let x b be the depth to the NA at balanced failure. From the strain relation,

yd

b

cu

b xdx

Þ ydcu

cub

dx

*

If x < x b Þ Steel yields first

If x > x b Þ Crushing of concrete takes place first.

S FH = 0 Þ Ts = CC Þ As fyd = 0.8 xb b fcd

Substituting for xb and simplifying,yd

cd

ydcu

cub f

f*

*8.0

(a steel ratio for balanced case)

However, for ductility purpose the steel ratio ρ may range b/n 0.75 ρb to 0.9 ρb, and in

some cases as low as 0.5 ρ b in ACI code, but in EBCS-2 ductility is ensured by keeping

kx max = 0.448 for 0% redistribution or even less for redistribution > 0% .

Rewriting the force equilibrium

byfcd = As fyd Þ b * 0.8x fcd = ρbd fyd

Where

S Mc = 0 Þ Md = As fyd (d - 0.4x)

Substituting the value of x and simplifying

Md = 0.8 bd2 fcd kx (1-0.4 kx)

When the above equation is solved for kx,

Where c1 = 2.5/m, c2 = 0.32m2fcd, m=fyd/(0.8fcd) kx max = 0.448 for 0% redistribution.

The section capacity for single reinforcement case may be computed from S Mt = 0, when kx < kx max

Þ Mu = 0.8bx fcd (d-0.4x) x = kx max d

BY: SILENAT D. 23


= 0.8bd2 fcd kx max (1 -0.4 kx max)

According to EBCS-2/95, elastic moments of continuous beams and frames are redistributed using the following

reduction coefficient,

1) For continuous beams and rigid jointed braced frames with span/effective depth ratio not greater than 20,

Where x—is calculated at ultimate limit state

Based on the above equation, the limiting maximum neutral axis depth ratio used for proportioning of sections of

continuous beams and rigid jointed braced frames are obtained as follow:

For 30% redistribution of elastic moment,



For no reduction of elastic moment,

2) For other continuous beams and rigid braced frames

3) For sway frames with slenderness ratio l of columns less than 25

Examples on Design of Singly Reinforced Beams using Limit State Design Method

BY: SILENAT D. 24


BY: SILENAT D. 25


BY: SILENAT D. 26


BY: SILENAT D. 27


BY: SILENAT D. 28


BY: SILENAT D. 29


BY: SILENAT D. 30


BY: SILENAT D. 31


BY: SILENAT D. 32


BY: SILENAT D. 33


BY: SILENAT D. 34


BY: SILENAT D. 35


BY: SILENAT D. 36


BY: SILENAT D. 37


BY: SILENAT D. 38


BY: SILENAT D. 39


BY: SILENAT D. 40


BY: SILENAT D. 41


BY: SILENAT D. 42


BY: SILENAT D. 43


Exercise-1

BY: SILENAT D. 44

AS

a) x-section b) Strainc) stresses, doubly reinforced section,

d) Stresses, balanced section,

e) stresses, excess tension-steel plus comp. steel,


3.3. Doubly Reinforced Rectangular Section

Consider a doubly reinforced rectangular section subjected to an ultimate moment, as shown below. Design

equations are derived by dividing the section into two parts: Balanced singly reinforced section and excess tension steel

plus compression steel. It is assumed that both tension and compression steels are yielded. The excess tension steel and

compression steel are proportioned in such a way that the neutral axis is maintained at balanced position.

Let --moment capacity of balanced singly reinforced section

--moment resistance provided by excess tension steel plus compression steel

Thus, the total ultimate moment of resistance of doubly reinforced section is the sum of the two parts: moment capacity

of balanced singly reinforced section, and ultimate moment resisted by excess tension steel plus compressive steel,

.

i.e

Moment capacity of balanced singly reinforced section,

And, the corresponding area of tension steel balancing is,

Where

Excess moment to be resisted by excess tension steel plus compression steel is,

BY: SILENAT D. 45


Equating excess moment with the couple made by internal forces in excess tension steel and compression steel as shown

in Fig.(e), area of excess tension steel and compression steel are obtained as (if compression steel yielding)

And,

Therefore, the total area of tension steel required by doubly reinforced section,

To check yielding of compression steel, referring to stain diagram in Fig.(b), the strain in compression steel is

determined and compared with the yield strain of a given steel as obtained below.

Where

If compression steel is yielding,

& (as assumed)

Or, if compression steel is not yielding,

&

Then, area of compression steel is re-determined using,

BY: SILENAT D. 46


Another Similar approach:

Assume that As & As1 are stressed to fyd.

Mu = Muc+ Musc

Where Muc is the BM carried by the concrete and partial area of tensile steal.

Muc = 0.8bd2 fcd k1 (1-0.4 k1)

In which k1 = kx max, the maximum steel ratio corresponding to single reinforcement section in case of design

and

for the case of analysis.

Musc is the BM carried by compressive steel and the corresponding tensile steel.

Musc = As1 fyd (d-dc’)

The yielding of the compressive steel may be checked from the strain relation as

BY: SILENAT D. 47


Examples on Design of Doubly Reinforced Beams using Limit State Design Method

BY: SILENAT D. 48


BY: SILENAT D. 49


BY: SILENAT D. 50


BY: SILENAT D. 51


BY: SILENAT D. 52


BY: SILENAT D. 53


BY: SILENAT D. 54


BY: SILENAT D. 55


BY: SILENAT D. 56


BY: SILENAT D. 57


3.4. Flanged Section (T- or L-section) under Flexure

The general discussion with respect to flanged section, effective width of flange in working stress method holds for

strength limit state method as well. In treating flanged section using strength limit state method, it is convenient to adopt

the same equivalent rectangle stress-block that is used for rectangular cross section.

i) If depth of equivalent rectangle stress-block, is equal to or less than the flanged thickness, (i.e

), a flanged section may be treated as a rectangular section of width equal to an effective width of flange,

provided the flange of section is on compression side when the section subjected a moment.

For trial purpose initially, it can be assumed the stress block is with-in the flange (or assume flanged section

rectangular with width equal to effective width of flange).

-calculate relative ultimate moment and relative mechanical steel ratio of assumed rectangular section using,

And

-then, compute depth of equivalent rectangle stress-block for assumed section and compare with thickness of the flange

of the section,

-If , the section is designed as rectangular section with width equal to effective width of flange, .

Therefore, area of tension steel required by the section for such case is given by

Where

ii) If the depth of equivalent rectangle stress-block of assumed rectangular section is greater than thickness of the flange

of the section (i.e ), a flanged section is treated as T-beam section provided the flange of section is on

compression side when the section subjected a moment. To derive design equation of T-beam, it is convenient to

divide the compression area of T-beam section into two parts as shown below:

a) the over-hanging portion of the compressive flange

b) the web portion extending into the compressive flange

BY: SILENAT D. 58

x-section Strain StressesOver-hanging

portionWeb portion

extending into flange


Let --area of tension steel balancing over-hanging portion of the flange

--area of tension steel balancing web portion extending into the flange

The total ultimate moment of resistance of T-beam section is obtained by taking moment of the internal compressive

forces about the center of tension steel; and it is given as the sum of moments produced by over-hanging portion of the

flange and the web portion extending into the flange. i.e

-The moment produced by over-hanging portion of the flange is obtained as

Where

Then, the corresponding area of tension steel balancing the over-hanging portion of the flange is obtained as

-The moment produced by the web portion extending into the flange is obtained by subtracting moment of over-hanging

portion from the total ultimate moment of T-beam.

i.e

To determine the corresponding area of tension steel balancing web potion extending into the flange, the web portion is

treated as rectangular section with width equal to the width of the web, . Therefore, calculate the relative ultimate

moment the web portion using

Then, the required area of tension steel balancing web potion is obtained as

Where

Therefore, the total area of tension steel is obtained as

BY: SILENAT D. 59


Check flanged section for single reinforcement using . If the flanged section requires compression

reinforcement ( ), area of compressive steel and excess tension steel required by web portion is obtained using

(if compression steel is yielding)

and, area of tension steel balancing web portion is re-determined using

Where &

iii) If the flange of the section is on the tension side when subjected to a moment, flanged section is designed as if it

were a rectangular section with width equal to the width of the web, .

Another similar approach:

Reinforced concrete floors or roofs are monolithic and hence, a part of the slab will act with the upper part of

the beam to resist longitudinal compression. The resulting beam cross-section is, then, T-shaped (inverted L),

rather than rectangular with the slab forming the beam flange where as part of the beam projecting below the

slab forms the web or stem.

The T-sections provide a large concrete cross-sectional area of the flange to resist the compressive force.

Hence, T-sections are very advantageous in simply supported spans to resist large positive bending moment,

where as the inverted T-sections have the added advantage in cantilever beam to resist negative moment.

BY: SILENAT D. 60

b

Dhf

be

Fig. 3.3.1


As the longitudinal compressive stress varies across the flange width of same level, it is convenient in design to

make use of an effective flange width (may be smaller than the actual width) which is considered to be

uniformly stressed.

Effective flange width (according to EBCS 2, 1995)

For interior beams T-sections

For edge beams inverted L- sections

Where le – is the effective span length & bw is the width of the web.

There are three distribution type of flexural behavior of T-sections.

When the T-section is subjected to BM, and tension is produced on the flange portion, it is treated as a

rectangular section with b = bw.

When the T-section is subjected to +ve bending moment and the equivalent compressive stress block lies

within the flange as shown below (y < hf), the section can be analysed as rectangular with effective width

be.

This case is a case of under reinforced condition or large flange thickness, which can be confirmed first

by computing (with b = be, = As/(bed)) using relation established for rectangular beams and evaluate

the NA depth, x = md. Compare y = 0.8x with hf.

BY: SILENAT D. 61bw

y


When y > hf, the section acts as T-beam and hence analysis accounting the T-geometry becomes essential

which is shown in the figure below.

Cross-section Design and Analysis

Design

- Assuming b = be compute and x = kx d

i) If y = 0.8x < hf, section is rectangular as assumed.

Þ As =

ii) If y > hf Þ T beam analysis is required.

As = ASf + Asw = + ρwbwd in which,

Muf = (be-bw) hf fcd zf

BY: SILENAT D. 62

bw

xh f

d

d'

be c

s

Cc

Ts

0.8x

f cd

Cross section Strain Stress


Muw = Mu - Muf

iii) When the flange is on the tension side, then the cross- section is designed as if it were rectangular with b

= bw

Analysis:

, X = ρmd

i) If y = 0.8X<= hf Þ the section is analyzed as rectangular with b = be.

Mu = 0.8bed2fcd ρm (1-0.4 ρm)

ii) If y = 0.8X< hf Þ the section is analyzed as T-beam.

Muf = (be-bw) hf fcd zf , ASf = , Asw = As - ASf

ρw = Muw = 0.8bwd2fcd ρwm(1-0.4ρwm)

Mu = Muf +Muw

Examples on Design of T-Section Beams using Limit State Design Method

BY: SILENAT D. 63


BY: SILENAT D. 64


BY: SILENAT D. 65


BY: SILENAT D. 66


BY: SILENAT D. 67


BY: SILENAT D. 68


BY: SILENAT D. 69


BY: SILENAT D. 70


BY: SILENAT D. 71


BY: SILENAT D. 72


BY: SILENAT D. 73


Alternative method using design tables (singly reinforced Sections)

1-USING DESIGN TABLES

Derivation

Md = 0.8bd2fcd ρm(1-0.4 ρm)

Let

∑Mc = 0 )4.01(

1*

)4.0(d

xfd

M

xdf

MA

yd

d

yd

ds

Þ

Let

BY: SILENAT D. 74


From table 1a there are different Km values and the max. Value of Km for different moment redistribution

is given and represented by Km*.

If Km ≤ Km*, the section is singly reinforced.

If Km>Km*, it is doubly reinforced.

STEPS:

a) For Singly Reinforced Sections

1. Evaluate

2. Enter the general design Table No.1a using km and concrete grade.

3. Read ks from the same Table corresponding to steel grade and km.

4. Evaluate

b) For Doubly Reinforced Sections

1. This is so, when Km>Km*(is the value of Km shown shaded in general design table 1a ,

corresponding to the concrete grade)

2. compute Km/Km*

3. Read Ks & Ks* corresponding to Km/Km* & the steel grade from general design table 1a

4. Assume dc, (d2) & read ρ (correction factor) from the same table corresponding to Km/Km* & dc’/d.

5. Read ρ’ corresponding to dc’/d ,then

As = KsMd ρ/d As’ = Ks’Md ρ’/d

Note: - In all cases

- Md is in KN-m

- b “ “ m

- d “ “ m

2- USING DESIGN CHARTS

BY: SILENAT D. 75


Compute & Kx, max = 0.8(δ-0.44), where δ=1, 0.9, 0.8 & 0.7 for 0%, 10%,

20% & 30% moment redistribution.

Compare or Kx with the corresponding values of Kx,max

Where: = 0.143, 0.205, 0.252 & 0.295 for 30%, 20%, 10%, and 0% respectively.

If ≤ then the section is singly reinforced and As1:

As1 =

If > ,then the section is doubly reinforced and As1 ,As2:

As2 = - area of compression reinforcement,

Where: Mu, s* = fcd bd2 & is the value given above.

- is actual compressive stress on compression steel & is Es*εsc

As1 = -area of tensile reinforcement

Using read Z/d, X/d etc & compute As1 and As2.

BY: SILENAT D. 76


BY: SILENAT D. 77


Cover to Reinforcements

BY: SILENAT D. 78


The concrete cover is the distance between the outermost surface of reinforcement (usually stirrups) and

the nearest concrete surface.

The thickness of cover required depends both upon the exposure conditions and on the concrete quality.

To transmit bond forces safely, and to ensure adequate compaction, the concrete cover should never be

less than:

(a) or n ( 40mm), or

(b) ( + 5mm) or (n + 5mm) if dg > 32mm

Where = the diameter of the bar.

n = the equivalent diameter for a bundle.

dg = the largest nominal aggregate size.

Minimum cover

Type of exposure Mild Moderate Sever

Min. cover (mm) 15 25 50

Durability and control of crack width is related with finishing and provision of adequate cover to

reinforcement. Nominal cover for structural elements located in the interior of the building with dry

environment and mild condition is 15 mm, example slab; humid environment with moderate exposure is 25

mm, example beam; severe environment is 50 mm, example footing.

Spacing of Reinforcements

The clear horizontal and vertical distance between bars shall be at least equal to the largest of the

following values.

(a) 20 mm

(b) The diameter of the largest bar or effective diameter of the bundle

(c) The maximum size of the aggregate dg plus 5mm.

Where bars are positioned in separate horizontal layers, the bars in each layer should be located

vertically above each other and the space between the resulting columns of the bars should permit the

passage of an internal vibrator.

Effective Span Length

BY: SILENAT D. 79


The effective span of a simply supported member shall be taken as the lower of the following two

values:

(a) The distance between the center lines of supports.

(b) The clear distance between the faces of supports plus the effective depth.

The effective span of a continuous element shall normally be taken as the distance between the center

lines of the supports.

For a cantilever, the effective span is taken to be its length, measured from.

(a) The face of the supports, for an isolated, fixed ended cantilever.

(b) The center line of the support for a cantilever which forms the end of a continuous beam.

Deflection limits are assumed to be satisfied when the minimum effective depth for a particular member is

where fyk is equal to character strength of reinforcement, Le is the effective span (the shorter span in case of two way slab), is constant, a function of restraints given below).

Table – values of

Member Simple End span Interior span cantilever

Beams 20 24 28 10

Slabs:Span ratio 2:1 25 30 35 12

Span ratio 1:1 35 40 45 10

* For intermediate values – interpolation.

Preliminary Sizing of Beam Sections

Ideal values of span effective depth ratios, recommended in the ISE manual for the preliminary sizing of

reinforced concrete beams are given in table below.

Support conditions Cantilever Simple Support Continuous End spans

ISE manual 6 12 15 13.5

3.6. One-way RC Slabs

BY: SILENAT D. 80


A reinforced concrete slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or

nearly so. It is used to provide flat surfaces mainly for roofs and floors of buildings, parking lots, air fields,

roadways …etc. It may be supported by reinforced concrete beams (and is poured monolithically with such

beams), by masonry or reinforced concrete walls, by structural steel members, directly by columns, or

continuously by the ground.

Classification: - Beam supported slabs may be classified as:-

1. One-way slabs – main reinforcement in each element runs in one direction only. (Ly/Lx >2). There are two

types- one way solid slabs and one way ribbed slabs.

2. Two – way

slabs – main

reinforcement

runs in both direction where ratio of long to short span is less than two. (Ly/Lx < 2)

Others include flat slabs, flat plates, two way ribbed or grid slabs etc.

3.6.1. Analysis of one-way solid slabs

BY: SILENAT D. 81

Solid slab Ribbed slabBeams

Joists


They are considered as rectangular beams of comparatively large ratio of width to depth and ratio of longer

span to width (short span) is greater than two.

When Ll / Ls > 2, about 90% or more of the total load is carried by the short span, i.e., bending takes place in

the direction of the shorter span.

The analysis is than carried out by assuming a beam of unit width with a depth equal to the thickness of the

slab and span equal to the distance between supports (in the short direction). The strip may be analyzed in the

same way as singly reinforced rectangular sections.

Load per unit area on the slab would be the load per unit length on this imaginary beam of unit width.

As the loads being transmitted to the supporting beams, all reinforcement shall be placed at right angles to

these beams. However some additional bars may be placed in the other direction to carry temperature and

shrinkage stresses.

Generally the design consists of selecting a slab thickness for deflection requirement and flexural design is

carried out by considering the slab as series of rectangular beams side by side

Remark:-

The ratio of steel in a slab can be determined by dividing the sectional area of one bar by the area of concrete

between two successive bars, the latter area being the product of the depth to the center of the bars and the

distance between them, center to center.

Unless condition warrant some change, cover to reinforcement is 15 mm.

The following minimum slab thicknesses shall be adopted in design:

a) 60mm for slabs not exposed to concentrated loads (eg. Inaccessible roofs).

b) 80mm for slabs exposed mainly to distributed loads.

c) 100mm for slabs exposed to light moving concentrated loads (eg. slabs accessible to light moving

vehicles).

BY: SILENAT D. 82

1m width

Ll

Ls

Supporting beams / walls


d) 120mm for slabs exposed to heavy dynamic moving loads (eg. slabs accessible to heavy vehicles).

e) 150mm for slabs on point supports (eg. flat slabs).

Flexural reinforcements should fulfill the following minimum criteria:

a) The ratio of the secondary reinforcement to the main reinforcement shall be at least equal to 0.2.

b) The geometrical ratio of main reinforcement in a slab shall not be less than:

c) The spacing between main bars for slabs shall not exceed the smaller of 2h or 350mm.

d) The spacing between secondary bars (in a direction to the main bars) shall not exceed 400mm.

3.6.2. Analysis and Design of one way Ribbed Slab

In one way ribbed slab, the supporting beams called joists or ribs are closely spaced. The ribbed floor is formed

using temporary or permanent shuttering (formwork) while the hollow block floor is generally constructed

with blocks made of clay tile or with concrete containing a light weight aggregate. This type of floor is

economical for buildings where there are long spans and light or moderate live loads such as in hospitals and

apartment buildings.

General Requirements:

Minimum slab thickness

To ensure adequate stiffness against bending and torsion and to allow ribbed slabs to be treated as solid slabs

for the purpose of analysis, EBCS-2 recommends that the following restrictions on size are satisfied:

Ribs shall not be less than 70mm in width; and shall have a depth, excluding any topping of not more than 4

times the minimum width of the rib. The rib spacing shall not exceed 1.0m

Thickness of topping shall not be less than 50mm, nor less than the clear distance between ribs. In the

case of ribbed slabs incorporating permanent blocks, the lower limit of 50mm may be reduced to 40mm.

BY: SILENAT D. 83bc

dw

dfbw

Rib Spacing

Cleardistance

Fig. Ribbed slab


Minimum Reinforcement

The topping shall be provided with a reinforcement mesh providing in each direction a cross-sectional area

not less than 0.001 of the section of the slab.

The breadth of ribs may be governed by shear strength requirements. The method proposed in the ISE

manual for the estimation of rib breadths limits the shear stress in the rib to 0.6 N/mm2 for concretes with

characteristic cylinder strength of 25 N/mm2 or more. The required breadth is given by:

b =

Where V is the maximum shear force in Newton’s on the rib considered as simply supported and d is the

effective depth in millimeters. For characteristic cylinder strengths less than 25 MPa, the breadth should be

increased in proportion.

If the rib spacing exceeds 1.0m, the topping shall be designed as a slab resting on ribs, considering load

concentrations, if any.

The function of the flanges with the web shall be checked for longitudinal shear.

The ultimate limit state in longitudinal shear is governed either by the effect of inclined flange compression

(acting parallel to its middle plane) or by tension in the transverse reinforcement.

The longitudinal shear per unit length vsd, which may be obtained as a function of the applied transverse shear

Vsd :

(a) For flange in compression :

vsd =

(b) For flange in tension.

vsd =

BY: SILENAT D. 84


Where: Vsd – applied transverse shear.

Vsd - longitudinal shear per unit length

be – effective width of a T-section.

z - Internal lever arm.

As – area of the longitudinal steel in the effective flanges outside the projection of

Web into the slab.

Asw – area of the longitudinal steel inside the slab within the projection of the

web into the slab.

Resistance to longitudinal shear.

(a) Resistance to inclined compression per unit length vRd1

vRd1 = 0.25 fcd hf

Where : hf = total thickness of the flange.

(b) Resistance to diagonal tension per unit length vRd2

vRd2 = 0.50 fctd hf +

Where : Asf = area of transverse reinforcement per unit length , perpendicular to

the web-flange interface.

If, at the section with M = Mmax , the flange is subjected to a tensile force, the concrete contribution 0.50 fctd

hf ( in the above equation) should be neglected.

BY: SILENAT D. 85Fd,max

sf hf

Asf

M = Mmax

Fd,max

av (shear span)

M = 0

Figure Forces on ribbed slab


Because joists are closely spaced, thickness of slab (topping),

Unless calculation requires for rib spacing larger than 1m, toppings or slabs are provided with mesh

reinforcement of 0.001 bD in both directions for temperature and shrinkage problem.

Unless calculation requires, min reinforcement to be provided for joists includes two bars, where one is bent

near the support and the other straight.

Rib with bw > 70mm, and overall depth Dj < 4 bw, joist + tslab

Rib spacing is generally less than 1m.

In case of ribbed spacing larger than 1m, the topping (slab) need to be design as if supported on ribs. (i.e.

As one way solid slab between the ribs).

If the span of the ribs exceeds 6m, transverse ribs may be provided, as the thickness of the topping will be

larger.

The girder supporting the joist may be rectangular or T-beam with the flange thickness equal to the floor

thickness.

Procedure of Design of a floor system of ribbed Slab

1. Thickness of toppings and ribs assumed based on min requirement.

2. Loads may be computed on the basis of center line of the spacing of joists.

3. The joists are analyzed as regular continuous T-beams supported by girders.

BY: SILENAT D. 86


4. Shear reinforcement shall not be provided in the narrow web of joist thus a check for the section capacity

against shear is carried out. The shear capacity may be approximated as: 1.1 Vc of regular rectangular

sections.

5. Determine flexural reinforcement and consider min provision in the final solution.

6. Provide the topping or slab with reinforcement as per temp and shrinkage requirement.

7. Design the girder as a beam.

Examples on Design of One Way slabs and Continuous Beams using Limit State Design Method

BY: SILENAT D. 87


BY: SILENAT D. 88


BY: SILENAT D. 89


BY: SILENAT D. 90


BY: SILENAT D. 91


BY: SILENAT D. 92


BY: SILENAT D. 93


BY: SILENAT D. 94


BY: SILENAT D. 95


BY: SILENAT D. 96


BY: SILENAT D. 97


BY: SILENAT D. 98


BY: SILENAT D. 99


BY: SILENAT D. 100


BY: SILENAT D. 101


BY: SILENAT D. 102


BY: SILENAT D. 103


BY: SILENAT D. 104


BY: SILENAT D. 105


3.7. Serviceability limits states of deflection and crack width

It is important that member performance in normal service be satisfactory, when loads are those actually

expected to act i.e. when load factors are 1.0. This is not guaranteed simply by providing adequate strengths.

Service load deflections under full load may be excessively large or long-term deflections due to sustained

loads may cause damage .Tension cracks in beams may be wide enough to be visually disturbing or may even

permit serious corrosion of reinforcing bars. These and other questions such as vibration or fatigue, require

consideration

Serviceability studies are carried out based on elastic theory, with stresses in both concrete and steel assumed

to be proportional to strain. The concrete on the tension side of the neutral axis may be assumed uncracked,

partially cracked, or fully cracked depending on the loads and material strengths.

Reinforced concrete members carrying lateral loads respond to these loads by bending. The moment curvature

relationship for a segment of the simply supported reinforced concrete member of fig.3.7.1 (a) is illustrated in

fig.3.7.1 (c). It can be seen that the segment remains uncracked and has a large stiffness EIu, , until the moment

reaches the cracking moment, Mcr, (Point A) .When this happens, the member cracks and the stiffness at the

cracked section reduces to EIc.

As the load (and hence the moment) is increased further, more cracks occur and existing cracks increase in

size .Eventually ,the reinforcement yields at the point of maximum moment corresponding to point C on the

diagram. Above this point the member displays large increases in deflection for small increases in

moment .The service load range is between the origin and point C on the diagram and it is in this range that

deflections are checked and stresses calculated.

Consider a point B within the service load range. This curvature represents the instantaneous (short term)

curvature under an applied moment, M. If the moment is sustained, however, the curvature increases with time

to point D owing to the creep of the concrete. The curvature at this point is known as the long term or

sustained curvature. As deflection results, from curvature, there are both instantaneous and sustained

deflections which must be considered in the design of members with bending.

BY: SILENAT D. 106


c) Moment/ curvature plot for segment of part (b)

Fig.3.7.1 Moment / Curvature relationship for beam

segment

3.7.1. Deflections

The deflections which result from bending must be limited such that they do not adversely affect the function

and appearance of the member or the entire structure.

a) Limits on Deflections

The final deflection (including the effects of temperature, creep and shrinkage) of all horizontal members shall

not, in general, exceed the value.

Where: Le effective span

For roof or floor construction supporting or attached to nonstructural elements (e.g. partitions and finishes)

likely to be damaged by large deflections, that part of the deflection which occurs after the attachment of the

non-structural elements shall not exceed the value .

BY: SILENAT D. 107

Yield point of reinforcement


b) Calculation of Deflections

Effect of creep and shrinkage strains on the curvature, and there by on the deflection shall be considered.

Immediate deflections shall be computed by the usual elastic methods as the sum of the two parts and

given by Eqs. 1 and 2, but not more than given by eqs. 3

------ (1)

--------------- (2)

---------------------- (3)

------------------------------------ (4)

= deflection due to the theoretical cracking moment (Mcr) acting on the uncracked transformed section

=deflection due to the balance of the applied moment over and above the cracking value and

acting on a section with an equivalent stiffness of 75% of the cracked value. = deflection of

fully cracked section

As = area of the tension reinforcement

Ecm = short term elastic modulus (secant modulus) of the concrete

Ecm = 9.5 fck-mpa, Ecm-Gpa

Grade of

concrete

C15 C20 C25 C30 C40 C50 C60

Ecm 26 27 29 32 35 37 39

Es-modulus of elasticity of steel, Iu-moment of inertia of the uncraked transformed section

Mk-Maximum applied, moment at mid span due to sustained characteristic loads; for cantilevers it is the

moment at the face of the support

S- Section modulus, d-effective depth of the section,

X-neutral axis depth at the section of max. moment,

Z-internal lever arm at the section of max moment.

-deflection coefficient depending on the loading and support conditions.

(e.g =5/48 for simply supported span subjected to uniformly distributed load)

BY: SILENAT D. 108


Note: The value of X & Z may be determined for the service load condition using a modular ratio of 10, or

for the ultimate load condition.

Long term deflection of flexural members shall be obtained by multiplying the immediate deflection caused by

the sustained load considered, by the factor,

(2-1.2As’/As) ³0.6--------------- (5)

Where: As’-area of compression reinforcement, As-area of tension reinforcement.

3.7.2. Limits on cracking

Flexural cracks are inevitably formed in reinforced concrete members. For structures in aggressive

environments, corrosion is a problem and stringent limits are imposed on the width of cracks that are allowed

to develop. Environment in the interior of the building is usually non-sever, corrosion does not generally pose

a problem and limits on crack widths will be governed by their appearance.

a) Crack Formation

The max. tensile stresses in the concrete are calculated under the action of design loads appropriate to

a serviceability limit state and on the basis of the geometrical properties of the transformed uncracked

concrete X-section.

The calculated stresses shall not exceed the following values:

a) Flexure, ( ) b) direct(axial) tension ( )

Minimum flexural reinforcement in beams for the control of cracking is given by:

b) Crack widths

BY: SILENAT D. 109


Crack widths are calculated using the quasi permanent service load combination. Specifically crack widths can

be assumed not to exceed the limiting values if the limits on the bar spacing or bar diameter (Table 1) are

satisfied, and if min. areas of reinforcement, also specified are provided.

Table 1 Maximum size and spacing of high bond bars for control of cracking.

Steel stress* Max. bar spacing (mm) Max. bar diameter(mm)

160

200

240

280

320

360

400

450

300

250

200

150

100

50

-

-

32

25

20

16

12

10

8

6

*steel stresses are determined using quasi –permanent loads.

Table 2 Characteristic crack widths for concrete Members

Type of

exposure

Dry environment:

Interior of buildings

of normal

habitation or office

(mild)

Humid environment:

Interior

components(e.g.

laundries), exterior

components;

components in non-

aggressive soil and

/or water

(Moderate)

Sea water and/or

aggressive chemicals

environments completely

or partially submergeed in

seawater ,components in

saturated salt

air ,aggressive industrial

atmospheres

(sever)

Characteristic

crack

width,wk(mm)

0.4 0.2 0.1

In specific cases where a crack width Calculation is considered necessary

BY: SILENAT D. 110


Wk= Where: wk=characteristic crack width, =average final crack width

=mean strain in the tension steel allowing for tension stiffening and time dependent effects

=coefficient relating the average crack width to the design value

for sections in bending under applied loads.

The mean strain is simply the strain in the steel adjusted by the distribution factor,

, Where: fs-stress in the tension reinforcement, Es-elastic modulus of steel

=coefficient which accounts for the bond properties of the reinforcement

=1.0 for high bond bars (normally used or deformed) and 0.5 for plain bars

= coefficient which accounts for the duration of loading or of repeated loading

=1.0 for single short term loading & 0.5 for sustained loading or repeated loading

fs= stress in tension steel assuming a cracked section

fsr= stress in tension steel assuming a cracked section due to loading which causes initial cracking

The average final crack spacing in (mm) is calculated using the equation

Srm = 50 + 0.25 111

Where: = coefficient which accounts for the bond properties of the reinforcement: k1=0.8 for

high bond bars:k1=1.6 for plain bars.

K2= coefficient which takes account of the form of strain distribution for bending it is 0.5

bar diameter, = effective reinforcement ratio As/Ac,eff.

Where: Ac,eff= effective tension area of he concrete , as illustrated below

BY: SILENAT D. 111


Exercise-2

BY: SILENAT D. 112

Effective tensionarea

Fig.3.7.2. Effective tension area of concrete

and (h-x)/3


BY: SILENAT D. 113


BY: SILENAT D. 114


BY: SILENAT D. 115


BY: SILENAT D. 116


BY: SILENAT D. 117


BY: SILENAT D. 118


BY: SILENAT D. 119


CHAPTER 4

THE ULS OF SHEAR AND BOND, ANCHORAGE AND DEVELOPMENT LENGTH

4.1 The ULS Design of Beams for Shear

Beams are designed for flexure and then the influences of other actions on its capacity are assumed.

The ULS of shear is characterized by either diagonal compression failure of concrete or failure of the web

reinforcement due to diagonal tension.

When a beam is subjected to flexure and shear, the shear resistance in the absence of shear reinforcement is

contributed by concrete compression zone, mechanical interlock of aggregate at the crack and dowel action

of the longitudinal reinforcement. The contributions of the later two are difficult to quantify.

Hence, the resistance to a diagonal tension is obtained as the sum of the resistance of the web reinforcement

and the concrete section.

In checking this resistance, the critical section for shear is assessed a distance d from the face of support.

4.2 Design Criteria

(i) Only nominal web reinforcement

When the shear force in a section does not exceed the shear strength of the concrete vc, only nominal web

reinforcement is provided.

Vc = 0.25 fctd K1 K2 bwd

Where: k1 = 1+50 ≤ 2.0

k2 = 1.6 – d ≥ 1.0, d is effective depth in m. For members where more than 50%

of the bottom reinforcement is curtailed, k2 = 1.0.

ρ = ≤ 0.02 (bw = the minimum width of the web)

BY: SILENAT D. 120


As = the area of the tensile reinforcement anchored beyond the intersection of

the steel and the line of possible 450 crack starting from the edge of the section.

When Vsd < Vc, the section is adequate and provide nominal web reinforcement specified by longitudinal

spacing as:

(a) All beams except joists of ribbed slabs, shall be provided with at least the minimum web reinforcement

given by:

Where: fyk is in MPa

Av = Pair area of stirrups

s = Spacing in mm

bw = width of web

(b) The maximum spacing smax between stirrups, in the longitudinal direction, shall be as given below.

smax = 0.5d ≤ 300mm if Vsd ≤

smax = 0.3d ≤ 200mm if Vsd >

(c) The transverse spacing of legs of stirrups shall not exceed d, or 800mm, which ever is the smaller.

(ii) Limiting value of ultimate shear stress

BY: SILENAT D. 121

dd

lb,net lb,net

Vsd Vsd

As As

Section considered

lb,net

Vsd

450

450

450


In order to prevent diagonal compression failure in the concrete the shear resistance (VRd) of a section shall

not be less than the applied shear force at d distance from face of support (Vsd).

Where, VRd = 0.25fcdbwd

When Vsd > VRd, the section size must be increased.

(iii) Shear reinforcement

When VC <VD <VRd, shear reinforcement need be provided.

; Av = pair area of reinforcement

When inclined bars are used,

Where: = the angle of inclination from the horizontal.

4.3. Bond, Anchorage and Development Length

BY: SILENAT D. 122

As AvnAvf yd


4.3.1. Bond

In order for reinforced concrete to behave as intended, it is essential that bond forces be developed on the

interface between concrete and steel, such as to prevent significant slip from occurring at that interface. If the

bar is smooth enough to slip, the assumption that the strain in an embedded reinforcing bar is the same as that

in the surrounding concrete, would not be valid. Consequently, the beam would be very little stronger than if it

were built of plain concrete, without reinforcement.

Formerly

plain bars were used with provision of end anchorage in the form of hooks. Such beam forms a broken bond

over the entire length between anchorages and acts as a tied arch (Fig. 4.3.2).

Figure 4.3.2 Tied arch action in a beam with little or no bond

BY: SILENAT D. 123

Figure 4.3.1 Bond stresses due to flexure (a) beam before loading; (b) unrestrained slip between concrete and steel; (c) bond forces acting on concrete; (d) bond forces acting on steel.

Figure 4.3.2 Tied arch action in a beam with little or no bond


To avoid development of wide cracks and dispense with special anchorage devices, deformed bars are now

universally used. With such bars, the shoulders of the projecting ribs bear on the surrounding concrete and

result in greatly increased bond strength.

4.3.2. Bond Stress

Figure 6.2.1 shows forces in an isolated piece of a beam of length dx. The moment at one end will generally

differ from that at the other end by a small amount dM.

Assuming that concrete does not resist any tension stresses, the change in bar force becomes,

(Z – Moment arm)

As shown in figure 4.3.1b, this force is resisted by the bond at the contact surface between bar and concrete.

Summing horizontal forces,

Where: u = local average bond stress per unit of bar surface area.

= sum of perimeters of all the bars.

Hence, the unit bond stress is proportional to the shear at a particular section, i.e., to the rate of change of

bending moment. The above equation applies to the tension bars in a concrete zone that is assumed to be fully

cracked. It does not apply to compression reinforcement, for which it can be shown that the flexural bond

stresses are very low.

BY: SILENAT D. 124

Figure 4.3.3 Forces and stresses acting on elemental length of beam: (a) Free body sketch of reinforced concrete element; (b) Free body sketch of steel element.


Actual distribution of flexural bond stress:

BY: SILENAT D. 125

Figure 4.3.4 variation of steel force and bond stress in reinforced concrete member subjected to pure bending: (a) cracked concrete segment; (b) bond stresses acting on reinforcing bar; (c) variation of tensile force in steel; (d) variation of bond stress along steel.


4.3.3 Development Length

Ultimate bond failures for bars in tension are of two types: the first is direct pullout of the bar, which occurs

when ample confinement is provided by the surrounding concrete. The second type of failure is splitting of the

concrete along the bar when cover, confinement or bar spacing is insufficient to resist the lateral concrete

tension resulting from the wedging effect of the bar deformations. The latter if more common than the former

BY: SILENAT D. 126

Figure 4.3.5 Effect of flexural cracks on bond stresses in beam (a) beam with flexural cracks; (c) variation of tensile force T in steel along span; (d) variation of bond stress u along span.


The development length is defined as that length of embedment necessary to develop the full tensile strength of

the bar, controlled by either pullout or splitting. Referring to figure 4.3.7, the moment, and hence the tensile

stress, is evidentially maximum at point a and zero at supports. The total tension force Abfs must be transferred

from the bar to the concrete in the distance l by bond stress on the surface.

The safety against bond failure is that the length of the bar, from any point of given steel stress ( fs or at most fy)

to its nearby free end must be at least equal to its development length.

The basic anchorage length, lb, is the straight length of bar required to anchor the force Asfyd. For a bar of

diameter-, this force must equal the shear force developed between the bar surface and the surrounding

concrete:

Where, fyd = design bond strength.

The required anchorage length lb,net depends on the type of anchorage and on the stress in the reinforcement and

can be calculated as:

BY: SILENAT D. 127

Figure 4.3.6 Splitting of concrete along reinforcement

Figure 4.3.7 Development length


Where, As,cal = theoretical area of reinforcement required by the design.

As,ef = area of reinforcement actually provided.

= 1.0 for straight bar anchorage in tension or compression.

0.7 for anchorage in tension with standard hooks.

For bars in tension, lb,min = 0.3lb ³ 10 or ³ 200mm

For bars in compression, lb,min = 0.6lb ³ 10 or ³ 200mm

Reinforcement shall extend beyond the point at which it is no longer required to resist tension for a length

given by:

(a) lb

(b) lb,net ³ d provided that in this case, the continuing bars are capable of resisting twice the applied moment

at the section.

BY: SILENAT D. 128

Figure 4.3.8 Standard Hooks


4.3.4 Bar Cut off and Bend points

It is a common practice either to cut off bars where they are no longer required to resist stress or in case of

continuous beams, to bend up bottom steel so that it provides tensile reinforcement at the top of the beam over

the support. To determine bend points, or bar cutting points, the moment diagram resulting from loading for

maximum span moment and maximum support moment is shown below.

Recognizing the various uncertainties, for bars with no special end anchorage the full development length lb,net

[d or 12] whichever is larger, must be provided beyond the peak stress location. The critical section may be

the point of max moment or a point where adjacent terminated reinforcement is no longer needed to resist

bending.

BY: SILENAT D. 129

Figure 4.3.9 Tensile force diagram


In the absence of explicit calculation, the sketch shown may serve this purpose.

BY: SILENAT D. 130

Figure 4.3.10 Curtailment of reinforcement – simply supported end


4.3.5 Bar splices

Reinforcing bars are as by fabrication limited in length, say 12 m. Thus it is normal to splice bars in the field.

To do this, one has to notice the following regarding splicing.

Splicing of bars must be avoided at points of max-moment.

Bars which are spliced should be staggered.

Splices are made simply by lapping the bars a sufficient distance to transfer stress

by bond from one bar to the other.

The required length of lap for tension is approximately 1.3lb and that for compression is lb.

BY: SILENAT D. 131

Figure 4.3.11 Simplified curtailment rules for beams


Examples on Design of Beams for Shear using Limit State Design Method

BY: SILENAT D. 132


BY: SILENAT D. 133


BY: SILENAT D. 134


BY: SILENAT D. 135


BY: SILENAT D. 136


BY: SILENAT D. 137


BY: SILENAT D. 138


BY: SILENAT D. 139


BY: SILENAT D. 140


Assignment-4

Question No. 3

Question No. 4

BY: SILENAT D. 141


Design of Typical Beam

Design of Continuous Beam on typical floor of a certain building

Depth determination

Section of the Beam

BY: SILENAT D. 142


D=450mm b=250mm fcd=11.33Mpa fyd=260.87Mpa Concrete Cover=25mm Reinf. Bar = Φ14mm

Stirrup=Φ8mm

Exterior Span,

d= (0.4+0.6

= 0.85 assume, 155mm

Overall depth, D = 155+25+7+8 = 195mm < Dused = 450mm

Interior Span,

d= (0.4+0.6

d = 0.85 assume, 145mm

Overall depth, D = 145+25+7+8 = 185mm < D used = 450mm

Design for Flexural Reinforcement

Calculate minimum reinforcement for beam according to EBCS 2, 1995, it is:-

ρmin= [EBCS 2,1995 Part 2 Section 7.2.1.1]

Asmin = ρbd = 0.002*250*410 = 205mm2

Design of Member T-10 left support

Moment = 119.97 KN-m

Critical moment at the face of Column = 0.9*119.97 = 107.97KN-m

μus =

0.01416 < 0.2952 , It is singly reinforced beam.

From chart, Kz = 0.98 Z = Kz * d = 0.98*0.41 = 0.4018

BY: SILENAT D. 143


As =

No. of bars assume 7 bars

Use 7Φ14 A s,provided = 1077.568mm2.

Member F-66 L,

y' =

d = 450-56.71 = 393.28mm.

μus =

kz = 0.85 , Z =0.3342 , As= 1238.15mm2, use 8Φ14 A s provided =1238.15mm2

BY: SILENAT D. 144


Member Support Moment(KN-m)

μus kz Z As

(mm2)No. of bars

A s provided

(mm2)

T-10 Left 107.974 0.01416 0.85 0.3342 1238.15 8Φ14 1231.49

Right 89.734 0.1884 0.87 0.3419 1066.05 7 Φ14 1077.568T-11 Left 41.458 0.0870 0.95 0.3895 408.01 3 Φ14 461.814

Right 52.613 0.1104 0.93 0.3813 528.934 4 Φ14 615.752T-12 Left 51.017 0.1071 0.94 0.3854 507.44 4 Φ14 615.752

Right 42.985 0.0902 0.94 0.3854 427.549 3 Φ14 461.814T-13 Left 86.143 0.1964 0.89 0.3649 904.952 6 Φ14 923.628

Right 111.488 0.2547 0.85 0.3342 1238.15 8 Φ14 1231.49

Design of positive moment on each Beam

Member T-10

Span moment 59.2017 KN-m

Check for T-beam

a) be = bw + b) be = C/C spacing between beams

= 250 + =5000mm

= 1250mm

Therefore, be = 1250mm governs.

M =

C1 =

C2 = 0.32* m2*fcd = 0.32*28.77*11.333*106 = 3002

Moment = 59.2017KN-m

BY: SILENAT D. 145


ρ=

= )

= 0.0013

ρ<ρmin → 0.0013<0.002, therefore, 0.002 governs.

X = ρ*m*d = 0.002*28.77*410 = 23.5914mm Y = 0.8X = 0.8*23.5914 = 18.873mm < hf =150mm. The beam is rectangular.

→ As = ρ*be*d = 0.002 (1250)(410) = 1025mm2

No. of bar =

Use 7Φ14 A s provided = 7(153.938) = 1077.567mm2

y' =

d = D-y'= 450-67.85 = 382.14mm.

Re-calculate As and check the new effective depth will change the reinforcement.

As = ρ*be*d = (0.002)*(1250)*(382.14) = 955.35mm2.

BY: SILENAT D. 146


No.of bars Re-calculate = 7 bars.

Use 7Φ14 A s provided =1077.567mm2.

Member

Moment(KN-m)

ρmin As

(mm2)No. of bars

A s provided

(mm2)T-10 59.2017 0.002 955.35 7Φ14 1077.567T-11 24.5233 0.002 820.00 6 Φ14 923.628T-12 24.5617 0.002 820.00 6 Φ14 923.628T-13 59.1943 0.002 955.35 7 Φ14 1077.567

Design for Shear

The minimum web reinforcement

ρmin = [EBCS 2, 1995 Sec.7.2.12]

The maximum spacing between stirrups

Smax = 0.5d < 300mm , if Vsd <2/3Vrd (EBCS 2,1995 Sec.3.5.4.1b)

Smax = 0.3d < 200mm , if Vsd>2/3 Vrd

Where :- Vsd = acting shear along critical section Vrd = shear resistance of section.

Vrd = 0.25 fcd *bw *d (EBCS 2, 1995 Sec.4.5.2.1)

The minimum shear resistance of concrete in beams

Vc = 0.25*fctd*k1*k2*bw*d (EBCS 2, 1995 Sec 4.5.3.1.1)

Where :- k1 = (1+50ρ) <2.0 k2= 1.6-d >1.0 (d in meters). For members where more than 50% of the bottom Reinforcement is contained, k2=1.

ρ= As/bd

Design of Shear Reinforcement

For longitudinal axis,

BY: SILENAT D. 147


Vs = (EBCS 2, 1995 Sec 4.5.4.4)

Where: - Av = is the area of shear resistance within the distance S.

Design for Shear Reinforcement

Span T-10

ρ=

k1 = (1+50ρ) <2.0 = (1+50*0.0125) <2.0 = 1.626 <2.0

k2 = (1.6-d) > 1.0 = (1.6-0.41) >1.0 = 1.19 >1.0

Vc = 0.25*fctd*k1*k2*bw*d = 0.25*(1.03*106)*1.626*1.19*0.25*0.41 = 47.918 KN.

Vrd = 0.25 * fcd * bw * d = 0.25 * (11.33*106) *0.25 * 0.41 = 290.408 KN.

Vsd right =

Vsd left =

BY: SILENAT D. 148


Vsd < 2/3Vrd , 105.215KN < 193.605KN , No diagonal compression failure.

Hence, Vsd > Vc , apply reinforcement to the web. Therefore, Smax = 0.5*410 = 205mm , Use Smax = 200mm.

Vs =

Vtot = Vs +Vc = 26.881 + 47.918 = 74.799KN

S = consider 50mm.

Hence, provide Φ8 c/c 200mm stirrup for the whole span.

Member Support ρ K1 Vc

(KN)Vrd

(KN)Vsd

(KN)Smax

(mm)Vs

(KN)Vtot

(KN)

T-10 Left 0.0105 1.525 47.88 290.41 104.2 Φ8c/c50mm 26.88 74.76Right 0.0105 1.525 47.88 290.41 100.1 Φ8c/c50mm 26.88 74.76




For mid span of each beam, we use minimum shear spacing ofUse T-10 Φ8c/c 50mm , T-11 Φ8c/c 90mm, T-12 Φ8c/c 90mm, T-13Φ8c/c 50mm

Development Length

Every reinforcement at each end must be anchored in order to avoid any bond failure.

BY: SILENAT D. 149


Basic anchorage length,

Type of reinforced bar deformed rough bar.

Lbnet =

=

fctk = 1.547Mpas .... for C-25fctd = 1.03 Mpasfbd = 2*fctd = 2*1.038 = 2.06Mpas.

Support T-10 Left

Ascal = 1238.15mm2. As.eff. = 1231.15mm2.

Lbnet =

Therefore, use Lbnet = 443.225mm

Member Support

As.cal

(mm2)As.eff.

(mm2)Lb.

(mm)Lb.net

(mm)T-10 Left 1238 1231.49 443.225 443.225

Right 942.66 1077.568 443.225 387.734T-11 Left 408.01 461.814 443.225 391.586

Right 528.934 615.752 443.225 380.732T-12 Left 507.44 615.752 443.225 365.261

Right 427.549 461.814 443.225 410.339T-13 Left 904.952 923.625 443.225 434.264

Right 1085.00 1231.49 443.225 390.501

Curtailment of Longitudinal Flexural Reinforcement

Longitudinal Flexural Reinforcement should extend up to or above the zero moment.

BY: SILENAT D. 150


Length of bar Support T-10 Left = 1.09+.44+0.2 = 1.74m. Support T-11 Left = 0.94+0.79+(2*.39) = 2.51m. Support T-11 Right & T-12 Left = 0.95+0.93+(2*.38) = 2.6m. Support T-12 Right & T-13 Left = 0.81+0.91+(2*0.43) = 2.59m. Support T-13 Right = 1.12+0.39+0.2 = 1.71mts.

BY: SILENAT D. 151

Documents

Rev. Rc i Chapter 1